Question

Which of the following aqueous solution has highest freezing point?

• A. \(0.1~mAl_{2}(SO_{4})_{3}\)
• B. \(0.1~m~BaCl_{2}\)
• C. \(0.1~m~NH_{4}Cl\)
• D. \(0.1~mAlCl_{3}\) \bigskip

Hint:

Highest freezing point \(\Rightarrow\) Smallest value of \(i\). For common electrolytes: \[ NH_4Cl(i=2) < BaCl_2(i=3) < AlCl_3(i=4) < Al_2(SO_4)_3(i=5) \]

Answer 1

The Correct Option is C

Concept: Freezing point depression is a colligative property and is given by \[ \Delta T_f=iK_fm \] where
• \(i\) = van't Hoff factor
• \(m\) = molality
• \(K_f\) = cryoscopic constant For solutions having the same molality, the solution with the smallest value of \(i\) undergoes the least freezing point depression and therefore possesses the highest freezing point.

Step 1: Calculate van't Hoff factor for each electrolyte.
For \(Al_2(SO_4)_3\), \[ Al_2(SO_4)_3 \rightarrow 2Al^{3+}+3SO_4^{2-} \] \[ i=5 \] For \(BaCl_2\), \[ BaCl_2 \rightarrow Ba^{2+}+2Cl^- \] \[ i=3 \] For \(NH_4Cl\), \[ NH_4Cl \rightarrow NH_4^+ + Cl^- \] \[ i=2 \] For \(AlCl_3\), \[ AlCl_3 \rightarrow Al^{3+}+3Cl^- \] \[ i=4 \]

Step 2: Compare the values of freezing point depression.
Since \[ \Delta T_f \propto i \] we obtain \[ Al_2(SO_4)_3 > AlCl_3 > BaCl_2 > NH_4Cl \] in terms of depression in freezing point.

Step 3: Determine the highest freezing point.
The smallest depression corresponds to \[ NH_4Cl \] Hence it has the highest freezing point. \[ \boxed{0.1m\ NH_4Cl} \]